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AP Calculus BC Flashcards: Fundamental Theorem Of Calculus Definite Intervals

Study Fundamental Theorem Of Calculus Definite Intervals in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Fundamental Theorem Of Calculus Definite Intervals, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Calculus BC Flashcards: Fundamental Theorem Of Calculus Definite Intervals

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QUESTION

What is the integral of f(x)=1f(x) = 1f(x)=1 from aaa to bbb?

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ANSWER

b−ab - ab−a. Antiderivative of 1 is xxx, difference of bounds gives length.

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Flashcard 1: What is the integral of f(x)=1f(x) = 1f(x)=1 from aaa to bbb?

Answer: b−ab - ab−a. Antiderivative of 1 is xxx, difference of bounds gives length.

Flashcard 2: Find the integral of f(x)=2x2−3x+1f(x) = 2x^2 - 3x + 1f(x)=2x2−3x+1 from 1 to 2.

Answer: 73\frac{7}{3}37​. Evaluate 2x33−3x22+x\frac{2x^3}{3} - \frac{3x^2}{2} + x32x3​−23x2​+x at bounds.

Flashcard 3: What is the definite integral of f(x)=5f(x) = 5f(x)=5 from 0 to 3?

Answer:

  1. Integral of constant function equals constant times interval length.

Flashcard 4: Find the integral of f(x)=3x2+2xf(x) = 3x^2 + 2xf(x)=3x2+2x from 0 to 2.

Answer:

  1. Evaluate x3+x2x^3 + x^2x3+x2 at bounds 2 and 0.

Flashcard 5: What is the integral of f(x)=e2xf(x) = e^{2x}f(x)=e2x from 0 to 1?

Answer: e2−12\frac{e^2 - 1}{2}2e2−1​. Antiderivative is e2x2\frac{e^{2x}}{2}2e2x​, evaluate at bounds.

Flashcard 6: What is the antiderivative of f(x)=2x3f(x) = 2x^3f(x)=2x3?

Answer: x42+C\frac{x^4}{2} + C2x4​+C. Power rule integration: increase exponent, divide by new exponent.

Flashcard 7: What does the Fundamental Theorem of Calculus connect?

Answer: Differentiation and integration. Links the inverse operations of calculus.

Flashcard 8: Find the integral of f(x)=1+xf(x) = 1 + xf(x)=1+x from 0 to 2.

Answer:

  1. Antiderivative is x+x22x + \frac{x^2}{2}x+2x2​, evaluate at bounds.

Flashcard 9: Calculate the integral of f(x)=x−1f(x) = x^{-1}f(x)=x−1 from 1 to 4.

Answer: ln(4)\text{ln}(4)ln(4). Same as ∫1xdx\int \frac{1}{x} dx∫x1​dx, antiderivative is ln⁡(x)\ln(x)ln(x).

Flashcard 10: Determine the integral of f(x)=exf(x) = e^xf(x)=ex from 0 to 1.

Answer: e−1e - 1e−1. Antiderivative of exe^xex is exe^xex, then e−1e - 1e−1.

Flashcard 11: Evaluate the integral of f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ from 1 to 2.

Answer: ln(2)\text{ln}(2)ln(2). Antiderivative of 1x\frac{1}{x}x1​ is ln⁡(x)\ln(x)ln(x), then ln⁡(2)−ln⁡(1)\ln(2) - \ln(1)ln(2)−ln(1).

Flashcard 12: Find the integral of f(x)=4x3−x2+2f(x) = 4x^3 - x^2 + 2f(x)=4x3−x2+2 from 0 to 1.

Answer: 134\frac{13}{4}413​. Evaluate x4−x33+2xx^4 - \frac{x^3}{3} + 2xx4−3x3​+2x at bounds 1 and 0.

Flashcard 13: Calculate the integral of f(x)=1x2f(x) = \frac{1}{x^2}f(x)=x21​ from 1 to 2.

Answer: 12\frac{1}{2}21​. Antiderivative is −1x-\frac{1}{x}−x1​, then −12−(−1)=12-\frac{1}{2} - (-1) = \frac{1}{2}−21​−(−1)=21​.

Flashcard 14: Evaluate the integral of f(x)=x3−3x2+2xf(x) = x^3 - 3x^2 + 2xf(x)=x3−3x2+2x from 0 to 1.

Answer:

  1. Evaluate x44−x3+x2\frac{x^4}{4} - x^3 + x^24x4​−x3+x2 at bounds 1 and 0.

Flashcard 15: Determine the integral of f(x)=x4f(x) = x^4f(x)=x4 from 0 to 1.

Answer: 15\frac{1}{5}51​. Antiderivative is x55\frac{x^5}{5}5x5​, then 15−0\frac{1}{5} - 051​−0.

Flashcard 16: What is the integral of f(x)=4xf(x) = 4xf(x)=4x from 1 to 3?

Answer:

  1. Antiderivative is 2x22x^22x2, then 18−2=1618 - 2 = 1618−2=16.

Flashcard 17: Find the value of the integral of f(x)=x2−xf(x) = x^2 - xf(x)=x2−x from 0 to 3.

Answer: 92\frac{9}{2}29​. Evaluate x33−x22\frac{x^3}{3} - \frac{x^2}{2}3x3​−2x2​ at bounds 3 and 0.

Flashcard 18: Calculate the integral of f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ from 1 to eee.

Answer:

  1. Antiderivative of 1x\frac{1}{x}x1​ is ln⁡(x)\ln(x)ln(x), then ln⁡(e)−ln⁡(1)=1\ln(e) - \ln(1) = 1ln(e)−ln(1)=1.

Flashcard 19: Identify the antiderivative given f(x)=2xf(x) = 2xf(x)=2x.

Answer: F(x)=x2+CF(x) = x^2 + CF(x)=x2+C. The antiderivative of 2x2x2x is x2x^2x2 plus a constant.

Flashcard 20: Evaluate the integral of f(x)=x3f(x) = x^3f(x)=x3 from 0 to 2.

Answer:

  1. Antiderivative is x44\frac{x^4}{4}4x4​, then 164−0=4\frac{16}{4} - 0 = 4416​−0=4.

Flashcard 21: What is the antiderivative of f(x)=2x3f(x) = 2x^3f(x)=2x3?

Answer: x42+C\frac{x^4}{2} + C2x4​+C. Power rule integration: increase exponent, divide by new exponent.

Flashcard 22: What does the Fundamental Theorem of Calculus connect?

Answer: Differentiation and integration. Links the inverse operations of calculus.

Flashcard 23: What is the integral of f(x)=1f(x) = 1f(x)=1 from aaa to bbb?

Answer: b−ab - ab−a. Antiderivative of 1 is xxx, difference of bounds gives length.

Flashcard 24: Determine the integral of f(x)=exf(x) = e^xf(x)=ex from 0 to 1.

Answer: e−1e - 1e−1. Antiderivative of exe^xex is exe^xex, then e−1e - 1e−1.

Flashcard 25: Evaluate the integral of f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ from 1 to 2.

Answer: ln(2)\text{ln}(2)ln(2). Antiderivative of 1x\frac{1}{x}x1​ is ln⁡(x)\ln(x)ln(x), then ln⁡(2)−ln⁡(1)\ln(2) - \ln(1)ln(2)−ln(1).

Flashcard 26: Find the integral of f(x)=4x3−x2+2f(x) = 4x^3 - x^2 + 2f(x)=4x3−x2+2 from 0 to 1.

Answer: 134\frac{13}{4}413​. Evaluate x4−x33+2xx^4 - \frac{x^3}{3} + 2xx4−3x3​+2x at bounds 1 and 0.

Flashcard 27: Identify the antiderivative given f(x)=2xf(x) = 2xf(x)=2x.

Answer: F(x)=x2+CF(x) = x^2 + CF(x)=x2+C. The antiderivative of 2x2x2x is x2x^2x2 plus a constant.

Flashcard 28: Calculate the integral of f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ from 1 to eee.

Answer:

  1. Antiderivative of 1x\frac{1}{x}x1​ is ln⁡(x)\ln(x)ln(x), then ln⁡(e)−ln⁡(1)=1\ln(e) - \ln(1) = 1ln(e)−ln(1)=1.

Flashcard 29: Find the value of the integral of f(x)=x2−xf(x) = x^2 - xf(x)=x2−x from 0 to 3.

Answer: 92\frac{9}{2}29​. Evaluate x33−x22\frac{x^3}{3} - \frac{x^2}{2}3x3​−2x2​ at bounds 3 and 0.

Flashcard 30: What is the integral of f(x)=4xf(x) = 4xf(x)=4x from 1 to 3?

Answer:

  1. Antiderivative is 2x22x^22x2, then 18−2=1618 - 2 = 1618−2=16.